The area of the triangle enclosed by the stra | vedclass

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(B) The given lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $x + 2y + 3 = 0$.To find the vertices of the triangle,we find the intersection points:$1

Vedclass · Accepted Answer

$2.25$

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(B) The given lines are $x = 0$ (y-axis),$y = 0$ (x-axis),and $x + 2y + 3 = 0$.To find the vertices of the triangle,we find the intersection points:$1$. Intersection of $x = 0$ and $y = 0$ is $(0, 0)$.$2$. Intersection of $x = 0$ and $x + 2y + 3 = 0$: Substituting $x = 0$,we get $2y = -3$,so $y = -1.5$. The point is $(0, -1.5)$.$3$. Intersection of $y = 0$ and $x + 2y + 3 = 0$: Substituting $y = 0$,we get $x = -3$. The point is $(-3, 0)$.The triangle is a right-angled triangle with base length $|-3| = 3$ and height $|-1.5| = 1.5$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 3 	imes 1.5 = \frac{4.5}{2} = 2.25$ sq. units.

The area of the triangle enclosed by the straight lines $x = 0$,$y = 0$ and $x + 2y + 3 = 0$ in sq. units is:

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